Question

How do you solve `abs(3x-4)<=x`?

Answer 1

`|3x-4|<=x => x in [1,2]`

or `1<=x<=2`

Explanation:

2 cases

`3x-4` is positive

`|3x-4|>=0 => 3x-4=3x-4<=x`

or

`3x-4` is negative

`|3x-4|<0 => |3x-4|=4-3x<=x`

If `3x-4` is positive then,

`3x-4<=x`

`<=>`

`3x<=x+4` Add 4 to both sides

`<=>`

`3x-x=2x<=4` Subtract x from both sides

`<=>`

`x<=2` Divide both sides by 2

Or `x in (-oo,2]`

If `3x-4` is negative then,

`4-3x<=x`

`<=>`

`4<=4x` Add `3x` to both sides

`<=>`

`1<=x` Divide both sides by 4

Or `x in[1,oo)`

Notice that

`(-oo,2]nn[1,oo)=[1,2]`

Then

`x in [1,2]`

Answer 2

`x in [1, 2]`

Explanation:

Here's one method...

Given:

`abs(3x-4) <= x`

Note in passing that `abs(...) >= 0` and hence `x >= 0`.

Given that `x >= 0`, we can square both sides of the inequality to get:

`9x^2-24x+16 <= x^2`

Subtract `x^2` from both sides to get:

`8x^2-24x+16 <= 0`

Divide both sides by `8` to get:

`x^2-3x+2 <= 0`

Factorise the quadratic to find:

`(x-1)(x-2) <= 0`

So this is a parabola with positive `x^2` coefficient, intersecting the `x` axis at `(1, 0)` and `(2, 0)`

Hence the solution of our inequality is:

`1 <= x <= 2`

In interval notation:

`x in [1, 2]`

Answer 3

`x in [1, 2]`. See the segment of of the x-axisl in the Socratic graph. for this solution.

Explanation:

graph{(|3x-4|-x-y)<=0 [-5, 5, -2.5, 2.5]}

`|3x-4|<=x` is the combined inequality for

`3x-4<=x`, giving `x<=2`, when `x>=4/3` and

`-(3x-4)<=x`, giving `x>=1`, when `x<=4/3`.